A uniform electric field of magnitude 336 N/C is directed along the +y-axis. A 5.30 µC charge moves from the origin to the point (x, y) = (-10 cm, -37 cm).
(a) What is the change in the potential energy associated with this charge?
To find the change in potential energy associated with the charge, we need to use the formula:
ΔPE = q * ΔV
where ΔPE is the change in potential energy, q is the charge, and ΔV is the change in voltage or potential.
In this case, we have a uniform electric field along the +y-axis, so the change in potential can be calculated using the formula:
ΔV = -E * d
where ΔV is the change in potential, E is the magnitude of the electric field, and d is the distance moved in the direction of the electric field.
Given:
E = 336 N/C (magnitude of the electric field)
q = 5.30 µC (charge)
d = -37 cm (distance moved in the direction of the electric field)
First, we need to convert the distance from centimeters to meters:
d = -37 cm = -0.37 m
Now we can calculate the change in potential:
ΔV = -E * d
ΔV = -(336 N/C) * (-0.37 m)
ΔV = 124.32 J/C
Finally, we can calculate the change in potential energy:
ΔPE = q * ΔV
ΔPE = 5.30 µC * 124.32 J/C
To find the final answer, we need to convert the charge from microcoulombs to coulombs:
ΔPE = (5.30 * 10^-6 C) * 124.32 J/C
Calculating this expression will give us the change in potential energy associated with the charge.